Theorem 1 (Mean and variance in an exponential dispersion family) Let \(Y\) have the density \(f(y; \theta, \phi)\) of Definition 11, and suppose that the support of \(Y\) does not depend on \(\theta\) and that \(b\) is twice differentiable. Then
\[
\begin{aligned}
{\mathbb{E}\mathopen{}\left[Y\right]\mathclose{}} &= b'(\theta)
\\
{\operatorname{Var}\mathopen{}\left(Y\right)\mathclose{}} &= \phi\, b''(\theta)
\end{aligned}
\]
[9, Ch. 5], [10, Ch. 3].
Proof. We write the proof with integrals; for a probability mass function, replace each integral with a sum. Because the support does not depend on \(\theta\), we can differentiate under the integral sign.
The derivative of the density with respect to \(\theta\) is
\[
\begin{aligned}
\frac{\partial}{\partial \theta} f(y; \theta, \phi)
&= f(y; \theta, \phi)
\frac{\partial}{\partial \theta}\mathopen{}\left[\frac{y\theta - b(\theta)}{\phi} + c(y, \phi)\right]\mathclose{}
&& \text{(chain rule for the exponential)}
\\ &= f(y; \theta, \phi) \frac{y - b'(\theta)}{\phi}
&& \text{(differentiate term by term)}
\end{aligned}
\]
Every density integrates to 1, for every value of \(\theta\), so differentiating that identity once gives the mean:
\[
\begin{aligned}
0 &= \frac{\partial}{\partial \theta} \int f(y; \theta, \phi)\, dy
&& \text{(derivative of the constant 1)}
\\ &= \int \frac{\partial}{\partial \theta} f(y; \theta, \phi)\, dy
&& \text{(differentiate under the integral)}
\\ &= \int \frac{y - b'(\theta)}{\phi} f(y; \theta, \phi)\, dy
&& \text{(derivative of the density)}
\\ &= \frac{{\mathbb{E}\mathopen{}\left[Y\right]\mathclose{}} - b'(\theta)}{\phi}
&& \text{(definition of expectation, and } \textstyle\int f = 1\text{)}
\end{aligned}
\]
Multiplying both sides by \(\phi > 0\) gives \({\mathbb{E}\mathopen{}\left[Y\right]\mathclose{}} = b'(\theta)\).
Differentiating the same identity a second time gives the variance:
\[
\begin{aligned}
0 &= \frac{\partial}{\partial \theta} \int \frac{y - b'(\theta)}{\phi} f(y; \theta, \phi)\, dy
&& \text{(derivative of the constant 0)}
\\ &= \int \mathopen{}\left[-\frac{b''(\theta)}{\phi} f(y; \theta, \phi)
+ \frac{y - b'(\theta)}{\phi} \frac{\partial}{\partial \theta} f(y; \theta, \phi)\right]\mathclose{} dy
&& \text{(product rule)}
\\ &= \int \mathopen{}\left[-\frac{b''(\theta)}{\phi}
+ \frac{\mathopen{}\left(y - b'(\theta)\right)\mathclose{}^2}{\phi^2}\right]\mathclose{} f(y; \theta, \phi)\, dy
&& \text{(derivative of the density)}
\\ &= -\frac{b''(\theta)}{\phi} + \frac{{\mathbb{E}\mathopen{}\left[\mathopen{}\left(Y - b'(\theta)\right)\mathclose{}^2\right]\mathclose{}}}{\phi^2}
&& \text{(definition of expectation, and } \textstyle\int f = 1\text{)}
\\ &= -\frac{b''(\theta)}{\phi} + \frac{{\operatorname{Var}\mathopen{}\left(Y\right)\mathclose{}}}{\phi^2}
&& \text{(} b'(\theta) = {\mathbb{E}\mathopen{}\left[Y\right]\mathclose{}}\text{, so this is the variance)}
\end{aligned}
\]
Multiplying both sides by \(\phi^2\) and rearranging gives \({\operatorname{Var}\mathopen{}\left(Y\right)\mathclose{}} = \phi\, b''(\theta)\).